Question

The weight distribution of parcels sent in a certain manner is normal with mean of 12 lb and standard deviation of 3 lb. The parcel service wishes to establish a weight value c beyond which there will be a surcharge. What value of c is such that 99% of all parcels are under the surcharge weight?

Answer 1

21.16

Explanation:

Starting from the theory we have the following equation:

`fi*P(x<c-1) = 0.99`

Using the data supplied in the exercise, we have subtracting the mean and dividing by the standard deviation:

`P( z \leq (c-1-12)/(3.5)) =0.99/fi`

solving for "c", knowing that fi is a tabulating value:

`(c-13)/(3.5)=0.99/fi\\(c-13)/(3.5)=2.33\\c-13=2.33*3.5\\c = 8.155 +13\\c = 21.155`

therefore the value of c is equal to 21.16